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NDA Math Indefinite Integrals

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For the next two (02) items that follow:

Consider

\int x \tan^{-1} x \, dx = A(x^2+1) \tan^{-1} x + Bx + C

where C is the constant of integration.

[NDA II 2014]

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Question : 38 of 38

Marks: +1, -0

What is the value of B?

1
1/2
-1/2
1/4

Solution:

\int x \tan^{-1}x\,dx = A(x^2+1)\tan^{-1}x + Bx + C

\Rightarrow \tan^{-1}x \int x \,dx - \int \frac{1}{1+x^2} \frac{x^2}{2} \,dx

\Rightarrow \frac{x^2}{2}\tan^{-1}x - \frac{1}{2} \int \frac{x^2}{1+x^2} \,dx

\Rightarrow \frac{x^2}{2}\tan^{-1}x - \frac{1}{2} \int \left(1 - \frac{1}{1+x^2}\right) dx + C

\Rightarrow \frac{x^2}{2}\tan^{-1}x - \frac{x}{2} + \frac{1}{2}\tan^{-1}x + C

\Rightarrow \frac{1}{2}\bigl[\tan^{-1}x(x^2+1) - x\bigr] + C

\Rightarrow A = \frac{1}{2}

and

B = -\frac{1}{2}

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